# Infinitesimal rotation of classical fields: why are rotation group representations important?

I understand that $SO(3)$ representations are important in quantum physics, because eigenspaces of the Hamiltonian are irreps of $SO(3)$ if it is part of the symmetry group. But I don't see the reason why the irreps should be essential for classical field theories.

My professor talked about the conservation of angular momentum because of Noether's theorem. The infinitesimal symmetry is

$$x^\mu \mapsto x^\mu + \omega^{\mu\nu}x_\nu$$

with an asymmetric matrix $\omega$, so the field transforms as

$$\phi_i(x) \mapsto {S_i}^j(\omega)\phi_j(x^\mu + \omega^{\mu\nu}x_\nu)$$

where $S$ is the representation of $SO(3)$, which is trivial for scalar fields.

• Why are representations necessary for such an infinitesimal symmetry? For translations, the $x^\mu$ transformation was sufficient, no talk about representations.
• If the trivial irrep is for scalars, are all the other (real) irreps for vectors, and their tensor products for higher tensors?

## 1 Answer

No one talked about representations for the translations because the translations do not change the natural basis for the vectors. When you apply a rotation, you are rotating the natural coordinate system your vectors and tensors are represented in, so fields which are components of those have to change. Translations don't change the natural coordinate basis at a point, so they don't affect such fields. In other words, no one talked about representations of the translation group because every field you normally think of transforms in the trivial representation.

The lowest non-trivial irreducible representation (labelled as "spin-1" in the usual diction) of $\mathrm{SO}(3)$, which is its fundamental representation, is the vector representation (let's call its representation space $V$). The tensor representations of rank $(k,l)$ are now $V\otimes\dots\otimes V\otimes V^*\otimes\dots\otimes V^*$ with $k$ factors of $V$ and $l$ factors of $V^*$. They are not irreducible, and split at least into the trace, symmetric traceless tensor and antisymmetric tensor representations.

The irreducible representations of higher spin do not usually play a special role, but they are not for "vectors"! For instance, the "next" representation labelled as "spin-2" is five-dimensional, and is that of the symmetric traceless matrices/rank-2 tensors.

• Thanks! Quick and very helpful. So the index in $\phi_i$ in the second formula does not necessarily run through $\{1,2,3\}$, it could also run through $\{1\}$ (for a scalar) or through cartesian products of $\{1,2,3\}$ for tensors? But for tensors, this would be a strange notation, wouldn't it? – Bass Sep 7 '15 at 19:33
• @BastianTreichler: Yes to all. Indeed, it conflicts somewhat with the usual notation for tensors, but I think whoever wrote this had primarily vector and spinor fields in mind, where there's no conflict. – ACuriousMind Sep 7 '15 at 19:45
• Please note my question on the $SO(3) irreducible representations on MSE: math.stackexchange.com/questions/1425825/… The third bullet point was inspired by your last paragraph. – Bass Sep 8 '15 at 14:30 • @BastianTreichler: You might be interested in this post of mine, where I show how to abstractly describe the irreducible representations from the algebra of ladder operators alone. Your first bullet point over there is incorrect, the higher spin representations do not act on$\mathbb{R}^3$, but on$\mathbb{R}^{2n+1}$(my ladder argument for the other post shows that there must be eigenvectors for every integer from$-n$to$n\$). – ACuriousMind Sep 8 '15 at 14:39